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Brut [27]
4 years ago
14

To determine drug dosages, doctors estimate a person’s body surface area (BSA) (in meters squared) using the formula BSA = √ h m

/ 60 , where h is the height in centimeters and m the mass in kilograms. Calculate the rate of change of BSA with respect to mass for a person of constant height h = 180 . What is this rate at m = 70 and m = 80 ? Express your result in the correct units. Does BSA increase more rapidly with respect to mass at lower or higher body mass?
Mathematics
1 answer:
krok68 [10]4 years ago
7 0

Answer:

Step-by-step explanation:

Given that a person’s body surface area (BSA) (in meters squared) using the formula

BSA =\frac{\sqrt{hm} }{60}

We are to calculate the rate of change of BSA with respect to mass for a person of constant height h = 180 from m =70 to m =80

For this we calculate BSA for 80 kg and BSA for 70 kg first

Rate of change = BSA(80)-BSA(70)

=\frac{\sqrt{180*80} }{60} -\frac{\sqrt{180*70} }{60} \\=2-1.871\\=0.129

i.e. rate of change per 10 kg is 0.129

No BSA does not increase more rapidly

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Answer with Step-by-step explanation:

Let

A=Students are female

B=Students major in civil engineering

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We have to find P(A/B)

P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{0.12}{0.19}=0.63

Hence, the probability that a randomly selected civil engineering major is female=0.63

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